It’s chock-full of amazing patterns (big surprise). I think I’ll spend a post or two (or three?) looking at a few of them.

First up: as noted by Steve Gilberg, adding up the numbers in each row reveals an interesting pattern: 1 = 1; 1+1 = 2; 1+2+1 = 4; 1+3+3+1 = 8; 1+4+6+4+1 = 16… amazing! The entries in each row add up to successive powers of two!

As a side note, let me tell you a story. Consider the polynomial function

Go substitute all the values from 1 to 6 into it for x, and tell me what you find. Things aren’t always what they seem.

OK, sorry, that wasn’t really a story, so sue me. Getting back to Pascal’s triangle, it *ahem* appears that the entries in each row add up to successive powers of two. At least, they do for the first eight or nine rows. But do they always? Simply observing this apparent pattern is a great start but it’s not enough.

I won’t keep you in suspense; it turns out that this is true. But can you figure out why? Post your explanations here! I know of at least two, one having to do with the way Pascal’s triangle is defined (in terms of the addition-rule), and one having to do with the way binomial coefficients are defined. But there are probably more.

25 Responses to More fun with Pascal's triangle (Challenge #9)

Well, the binomial coefficient proof is easy: Simply expand (1+1)^n, and you’ve got it. Incidentally, you can also use this to prove a bunch of other cool factoids. Try expanding 0 = (1-1)^n for starters.
But that’s not the coolest factoid, IMO. What can you prove about this sequence of numbers?

True, although that assumes you already know the binomial theorem. (I am thinking of writing a post on it soon.) There’s also a more intuitive way to explain it by appealing to the combinatorial meaning of binomial coefficients.

I took the liberty of surrounding your formula with [ tex ] and [ /tex ] (without the spaces) — anything in between those tags is interpreted as LaTeX, so no need to worry about HTML. At any rate, those factoids are indeed cool. Perhaps some other readers will take you up on your challenge!

No such thing as a silly question! (Well, I suppose that’s not technically true, but you know what I mean. =) How I got to that polynomial is probably deserving of a couple posts in and of itself, but I’ll see if I can sketch the method quickly. In general it is always possible to find a degree-n polynomial which will pass through any (n+1) points of your choosing (as long as no two points have the same x-coordinate). Say your polynomial is

Then saying you want it to pass through the point is the same as saying you want the equation to be true. Creating an equation out of each of your chosen points in this way gives you a system of n+1 linear equations (one for each of your points) with n+1 unknowns (a_n through a_0). You can then solve this system of equations to find the coefficients of the desired polynomial (in my case I had a computer do it for me, using matrices).

That’s probably long enough for one comment, but if you want more details than that, just let me know!

That’s right. I suppose I could have just made a degree-4 polynomial to pass through f(1)=1 to f(5)=16 and left f(6) up to chance (it probably would not be 32) but somehow I thought it would be more fun to have f(6) be so close to continuing the pattern… but not quite!

With this definition, it turns out that the the sum of the elements of any row is equal to twice the sum of the elements in the previous row. And since row 0 summed to 1, row 1 will sum to 2, row 2 will sum to 4, etc, all powers of 2.

2. Define (n, x) = nCx. So the triangle becomes

0C0
1C0 1C1
2C0 2C1 2C2
3C0 3C1 3C2 3C3
…etc…

Then once can prove the combinatorial identity that

nCx = (n-1)C(x-1) + (n-1)Cx

And then the argument follows in the same way as the previous way to define Pascal’s triangle.

I’ve got to stop by here more often (and I will, just added you to my links)

So I have a fun way, without reference to “that triangle”

Consider a set of n objects. How many subsets does it have? I have two solutions.

1. Consider the subsets of each different size (cardinality). There are C(n,n) (which is one) subset of order n, C(n,n-1) subsets of order n – 1, and so one, down to C(n,1) = n subsets of order 1 and of course C(n,0) 1 subset with nothing in it!

2. Lets count the subsets differently. Look at each element of the original set, and say “yes” we will include it, or “no” we will not. That’s 2 choices for the first element, 2 for the second, 2 for… Actually, it’s 2 choices for each. They are independent, so we multiply, and get 2^n

Since the left and right hand sides of your equation answer the same question, they must be equal.

This is way outside the purview of this particular post, but I thought I’d share my favorite Pascal property. The proof follows from the way in which the triangle is constructed and a combinatorial argument.

Pascal’s triangle describes the number of vertices, edges, faces, etc. of any n-dimensional triangle (a simplex, if you want to get technical about it).

Check out the 1-3-3-1 row. Disregard the first 1, so that you have 3-3-1.
How many vertices (zero-dimensional corners) are in a two-dimensional standard triangle? 3.
How many edges (one-dimensional sides) are in a triangle? 3.
How many faces (two-dimensional areas) are in a triangle? 1 — the triangle itself.

Next, check out the 1-4-6-4-1. Again, disregard the initial 1, so that you have 4-6-4-1.
How many vertices are in a three-dimensional triangle (aka tetrahedron, aka triangular pyramid)? 4.
How many edges are there in a tetrahedron? 6.
How many faces? 4.
How many spaces (three-dimensional volumes)? 1.

This generalizes up and down through all dimensions, though it obviously gets tricker to visualize. You can modify the rules of generating Pascal’s triangle so that it will do the same trick for n-dimensional squares, also.

Well, if you want an idea of how it is connected to the “real world”, you could note that binomial coefficients (the entries in Pascal’s triangle) tell you how many ways there are to choose k things out of n total things; then, the fact that each row sums to 2^n means that there are 2^n ways to choose as many (or as few) things as you want out of n things. For example, if an ice cream store has chocolate, vanilla, and strawberry, then you have 2^3 = 8 different choices (including not getting any ice cream, getting only strawberry, getting chocolate and vanilla, etc.).

But as for “what the pattern is for”, I’m not sure that’s the right question to ask. The pattern just is — if you write down a triangle of numbers where each one is the sum of the two above, and add up the rows, you get powers of two. The pattern exists because that’s the way the universe is structured, not because it is useful. =) Although it is pretty nifty that it happens to also correspond to a “real-world” interpretation.